274
7 Mathematical Models of Functionally Graded Beams in Temperature Field
Since σ yy = σ zz = 0, then using (7.102)–(7.104) and (7.111)–(7.116), the internal
energy U of the deformed isotropic three-layer beams takes the following form:
U = (1/2) b
L
0
c+h 1
−c−h 2
σ i j ε i j + m i j χ i j
dxdz =
= (1/2) b
L
0
⎡
⎣
c+h 1
c
E 1
∂u
∂ x
+ c
∂α
∂ x
− z
∂
2 w
∂ x 2
2
+ l
2
1 G 1
∂
2 w
∂ x 2
2
dz+
+
c
−c
E 3
∂u
∂ x
+ z
∂α
∂ x
− z
∂
2 w
∂ x 2
2
+ G 3 α
2
+
l
2
3 G 3
4
∂α
∂ x
− 2
∂
2 w
∂ x 2
2
dz+
+
−c
−c−h 2
E 2
∂u
∂ x
− c
∂α
∂ x
− z
∂
2 w
∂ x 2
2
+ l
2
2 G 2
∂
2 w
∂ x 2
2
dz
⎤
⎦ dx.
(7.142)
Let us introduce to the layers the virtual displacements δ u, δ w, δα owing to
(7.137)–(7.141). Therefore, the variation of the internal energy takes the form
δ U = b
L
0
⎡
⎣
c+h 1
c
σ xx
∂δu
∂ x
+ c
∂δα
∂ x
− z
∂
2
δw
∂ x 2
− m xy
∂
2
δw
∂ x 2
dz+
+
c
−c
σ xx
∂δu
∂ x
+ z
∂δα
∂ x
− z
∂
2
δw
∂ x 2
+ τ δα + m xy
1
2
∂δ α
∂ x
− 2
∂
2
δ w
∂ x 2
dz+
+
−c
−c−h 2
σ xx
∂δu
∂ x
− c
∂δα
∂ x
− z
∂
2
δw
∂ x 2
− m xy
∂
2
δw
∂ x 2
dz
⎤
⎦ dx.
(7.143)
Now, introducing the general displacement v instead if u, and taking into account
(7.126), we get
δu = δV −
1
2
h
c 12 δα − c 13
∂δw
∂ x
and hence (7.143) yields
δ U =
L
0
∂ N
∂ x
δV +
∂ H
∂ x
− Q 3
δα +
∂
2 M
∂ x 2 δw
dx−
−
N δV + M
γ δα −
∂δw
∂ x
+
H
γ
− M
γ δα +
∂ M
∂ x
δw
x=L
x=0
.
(7.144)